1961 AHSME Problems/Problem 27

Problem

Given two equiangular polygons $P_1$ and $P_2$ with different numbers of sides; each angle of $P_1$ is $x$ degrees and each angle of $P_2$ is $kx$ degrees, where $k$ is an integer greater than $1$ . The number of possibilities for the pair $(x, k)$ is:

$\textbf{(A)}\ \infty\qquad \textbf{(B)}\ \text{finite, but greater than 2}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$

Solution

Each angle in each equiangular polygon is $\frac{180(n-2)}{n} = 180 - \frac{360}{n}$ , where $n$ is the number of sides. As $n$ gets larger, each angle in the equiangular polygon approaches (but does not reach) $180^{\circ}$ . That means $kx < 180$ , so $x < \frac{180}{k}$ .

Recall that each angle in an equiangular triangle has $60^{\circ}$ , each angle in an equiangular quadrilateral has $90^{\circ}$ , and so on. If $k = 2$ , then $x < 90$ . That means the only option available is $x = 60$ because an equiangular triangle is the only equiangular polygon to have acute angles. If $k = 3$ , then $x < 60$ , and there are no values of $x$ that satisfy. As $k$ increases, $\frac{180}{k}$ decreases even further, so we can for sure say that there is only 1 possibility for the pair $(x, k)$ , which is answer choice $\boxed{\textbf{(D)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1961 AHSME Problems/Problem 27

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